Integer Inquiry

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 22451 Accepted Submission(s): 6099

Problem Description

One of the first users of BIT‘s new supercomputer was Chip Diller. He extended his exploration of powers of 3 to go from 0 to 333 and he explored taking various sums of those numbers.

``This supercomputer is great,‘‘ remarked Chip. ``I only wish Timothy were here to see these results.‘‘ (Chip moved to a new apartment, once one became available on the third floor of the Lemon Sky apartments on Third Street.)

Input

The input will consist of at most 100 lines of text, each of which contains a single VeryLongInteger. Each VeryLongInteger will be 100 or fewer characters in length, and will only contain digits (no VeryLongInteger will be negative).

The final input line will contain a single zero on a line by itself.

Output

Your program should output the sum of the VeryLongIntegers given in the input.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Sample Input

1

123456789012345678901234567890

123456789012345678901234567890

123456789012345678901234567890

0

Sample Output

370370367037037036703703703670

可以用string的高精度模板進行加法運算，
#include<iostream>
#include<string>
using namespace std;
string add(string str1,string str2)
{
    int len1=str1.length();
    int len2=str2.length();
    if(len1>len2)               //為了讓長度較小的字符串前面補0，使兩個字符串的長度相等。
    {
        for(int i=1;i<=len1-len2;i++)
            str2="0"+str2;
    }
    else
    {
        for(int i=1;i<=len2-len1;i++)
            str1="0"+str1;
    }
    len1=str1.length();
    int cf=0;
    int t;
    string str;
    for(int i=len1-1;i>=0;i--)             //因為最後一位數的下標就是len1-1;首位的為0。
    {
        t=str1[i]-‘0‘+str2[i]-‘0‘+cf;
        cf=t/10;                    //進位。
        t%=10;
        str=char(t+‘0‘)+str;
    }
    if(cf!=0) str=char(cf+‘0‘)+str;
    return str;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        string sum="0";
        string str1;
        while(cin>>str1)                //可以用這方法進行循環輸入字符串。
        {
            if(str1=="0")break;
            sum=add(sum,str1);
        }
        cout<<sum<<endl;
        if(t)
            cout<<endl;
    }
    return 0;
}

用java：

import java.math.BigInteger;
import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner in =new Scanner(System.in);
        int t=in.nextInt();
        while(t>0) {
            t--;
            BigInteger a=new BigInteger("0");
            while(in.hasNextBigInteger()) {
                BigInteger b=in.nextBigInteger();
                if(!b.equals(BigInteger.valueOf(0))) {
                    a=a.add(b);
                }
                else {
                    System.out.println(a);
                    if(t!=0)
                        System.out.println();
                    break;
                }
            }
        }
    }
}

（大數 string） Integer Inquiry hdu1047